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arXiv 1406.1016gr-qc
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比安奇恒等式总是双曲线的吗?

Is the Bianchi identity always hyperbolic?

István Rácz

摘要:

考虑满足爱因斯坦方程的$n+1$维光滑黎曼和洛伦兹空间.基流形被假设为光滑的单参数超曲面族。在这两种情况下-同样,它通常是在洛伦兹的情况下-爱因斯坦的方程可以分为“汉密尔顿”和“动量”的约束和“减少”的一套场方程。它表明,无论是主空间是黎曼或洛伦兹,每当叶状超曲面是黎曼的“哈密顿”和“动量”型表达式是附属的一阶对称双曲系统。由于这个辅助系统是线性的和齐次的“哈密尔顿”和“动量”型表达式的系统的双曲性意味着在这两种情况下的解决方案“减少”的一套场方程也是解决方案的全套方程提供的约束举行的超曲面之一foliating基地流形。

英文摘要:

We consider $n+1$ dimensional smooth Riemannian and Lorentzian spaces satisfying Einstein's equations. The base manifold is assumed to be smoothly foliated by a one-parameter family of hypersurfaces. In both cases---likewise it is usually done in the Lorentzian case---Einstein's equations may be split into `Hamiltonian' and `momentum' constraints and a `reduced' set of field equations. It is shown that regardless whether the primary space is Riemannian or Lorentzian whenever the foliating hypersurfaces are Riemannian the `Hamiltonian' and `momentum' type expressions are subject to a subsidiary first order symmetric hyperbolic system. Since this subsidiary system is linear and homogeneous in the `Hamiltonian' and `momentum' type expressions the hyperbolicity of the system implies that in both cases the solutions to the `reduced' set of field equations are also solutions to the full set of equations provided that the constraints hold on one of the hypersurfaces foliating the base manifold.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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